Question

Compute the derivative of the given function. $$f(t)=\sqrt[5]{t}\left(\sec t+e^{t}\right)$$

Short answer

The derivative is \( f'(t) = \frac{1}{5}t^{-4/5}(\sec t + e^t) + \sqrt[5]{t}(\sec t \tan t + e^t) \).

Step-by-step solution

Recognize the product rule requirement

The function is given as \( f(t) = \sqrt[5]{t} (\sec t + e^t) \). Identify that the function is comprised of two parts that are being multiplied together: \( u(t) = \sqrt[5]{t} \) and \( v(t) = \sec t + e^t \). The derivative of such a product requires the product rule, which states that if \( f(t) = u(t)v(t) \), then \( f'(t) = u'(t)v(t) + u(t)v'(t) \).

Differentiate \( u(t) = \sqrt[5]{t} \)

The expression \( u(t) = \sqrt[5]{t} \) can be written as \( t^{1/5} \). To differentiate this term with respect to \( t \), use the power rule: \( \frac{d}{dt}[t^n] = nt^{n-1} \). Therefore, \( u'(t) = \frac{1}{5}t^{-4/5} \).

Differentiate \( v(t) = \sec t + e^t \)

To differentiate \( v(t) = \sec t + e^t \), calculate the derivatives of each part separately. The derivative of \( \sec t \) is \( \sec t \tan t \), and the derivative of \( e^t \) is \( e^t \). Thus, \( v'(t) = \sec t \tan t + e^t \).

Apply the product rule

Use the product rule formula: \( f'(t) = u'(t)v(t) + u(t)v'(t) \). Substitute \( u'(t) = \frac{1}{5}t^{-4/5} \), \( v(t) = \sec t + e^t \), \( u(t) = \sqrt[5]{t} \), and \( v'(t) = \sec t \tan t + e^t \).

Calculate \( f'(t) \)

Calculate each component of the product rule:1. \( u'(t)v(t) = \frac{1}{5}t^{-4/5}(\sec t + e^t) \).2. \( u(t)v'(t) = \sqrt[5]{t}(\sec t \tan t + e^t) \).Combine these results to find \( f'(t) = \frac{1}{5}t^{-4/5}(\sec t + e^t) + \sqrt[5]{t}(\sec t \tan t + e^t) \).

Simplify the derivative

The expression for the derivative \( f'(t) = \frac{1}{5}t^{-4/5}(\sec t + e^t) + \sqrt[5]{t}(\sec t \tan t + e^t) \) can be left in this expanded form, as it simplifies only with additional factoring or grouping in specific contexts.

Key concepts

Product Rule

The product rule is essential when differentiating functions that multiply two separate expressions. If you have a function written as \( f(t) = u(t)v(t) \), the product rule is your go-to tool. It involves finding the derivative of both functions and combining them in a specific way: \[ f'(t) = u'(t)v(t) + u(t)v'(t) \] This means you'll differentiate the first function, multiply it by the second function as it is, and add to that the first function multiplied by the derivative of the second function.

• **Differentiate one function:** Find the derivative of the first function.
• **Multiply:** Multiply this derivative by the second function as it stands.
• **Differentiate the other function:** Find the derivative of the second function.
• **Combine:** Multiply this new derivative by the first function and add your results.

Power Rule

The power rule is straightforward and very useful when handling expressions that involve powers of a variable. Suppose you have a function \( t^n \), where \( n \) is any real number. The power rule allows you to differentiate it effectively. Here's how it works: \[ \frac{d}{dt}[t^n] = nt^{n-1} \] Simply bring down the power as a multiplier in front and decrease the power by one in the term. For instance, \( t^{1/5} \) turns into \( \frac{1}{5}t^{-4/5} \) when you apply the power rule. This rule is powerful for simplifying differentiation when variables are raised to a power.

Trigonometric Derivatives

Trigonometric functions have their own derivatives that are important when the functions involve sine, cosine, tangent, etc. Consider the derivative of the secant function, \( \sec t \), which is a bit more complex than simple constant derivatives.

• Secant Derivative: The derivative of \( \sec t \) is \( \sec t \tan t \), essential when differentiating \( \sec t \).

When you encounter a composite function like \( \sec t + e^t \), you need to handle each part separately using the appropriate rules. The trigonometric part, \( \sec t \), requires you to use its specific derivative formula.

Exponential Functions

Exponential functions are ubiquitous in calculus, often requiring differentiation. For a function like \( e^t \), the derivative is surprisingly simple and mirrors the function itself: \[ \frac{d}{dt}[e^t] = e^t \] This is one of the most elegant aspects of exponential functions, where the rate of change directly follows the function's form. When working with mixed functions like \( \sec t + e^t \), understanding exponential functions helps streamline the differentiation process.